8.7: Constant Coefficient Equations with Impulses
( \newcommand{\kernel}{\mathrm{null}\,}\)
So far in this chapter, we’ve considered initial value problems for the constant coefficient equation
ay″+by′+cy=f(t),
where f is continuous or piecewise continuous on [0,∞). In this section we consider initial value problems where f represents a force that’s very large for a short time and zero otherwise. We say that such forces are impulsive. Impulsive forces occur, for example, when two objects collide. Since it isn’t feasible to represent such forces as continuous or piecewise continuous functions, we must construct a different mathematical model to deal with them.
If f is an integrable function and f(t)=0 for t outside of the interval [t0,t0+h], then ∫t0+ht0f(t)dt is called the total impulse of f. We’re interested in the idealized situation where h is so small that the total impulse can be assumed to be applied instantaneously at t=t0. We say in this case that f is an impulse function. In particular, we denote by δ(t−t0) the impulse function with total impulse equal to one, applied at t=t0. (The impulse function δ(t) obtained by setting t0=0 is the Dirac δ function.) It must be understood, however, that δ(t−t0) isn’t a function in the standard sense, since our “definition” implies that δ(t−t0)=0 if t≠t0, while
∫t0t0δ(t−t0)dt=1.
From calculus we know that no function can have these properties; nevertheless, there’s a branch of mathematics known as the theory of distributions where the definition can be made rigorous. Since the theory of distributions is beyond the scope of this book, we’ll take an intuitive approach to impulse functions.
Our first task is to define what we mean by the solution of the initial value problem
ay″+by′+cy=δ(t−t0),y(0)=0,y′(0)=0,
where t0 is a fixed nonnegative number. The next theorem will motivate our definition.
Suppose t0≥0. For each positive number h, let yh be the solution of the initial value problem
ay″h+by′h+cyh=fh(t),yh(0)=0,y′h(0)=0,
where
fh(t)={0,0≤t<t0,1/h,t0≤t<t0+h,0,t≥t0+h,
so fh has unit total impulse equal to the area of the shaded rectangle in Figure 8.7.1 . Then
lim
where
w={\cal L}^{-1}\left(1\over as^2+bs+c\right).\nonumber
- Proof
-
Taking Laplace transforms in Equation \ref{eq:8.7.1} yields
(as^2+bs+c)Y_h(s)=F_h(s),\nonumber
so
Y_h(s)={F_h(s)\over as^2+bs+c}.\nonumber
The convolution theorem implies that
y_h(t)=\int_0^t w(t-\tau)f_h(\tau)\,d\tau.\nonumber
Therefore, Equation \ref{eq:8.7.2} implies that
\label{eq:8.7.4} y_{h}(t)=\left\{\begin{array}{cl}{0,}&{0\leq t< t_{0},}\\[4pt]{\frac{1}{h}\int_{t_{0}}^{t}w(t-\tau )d\tau , }&{t_{0}\leq t\leq t_{0} +h,}\\[4pt]{\frac{1}{h}\int_{t_{0}}^{t_{0}+h}w(t-\tau )d\tau , }&{t> t_{0}+h.} \end{array} \right.
Since y_h(t)=0 for all h if 0\le t\le t_0, it follows that
\label{eq:8.7.5} \lim_{h\to0+}y_h(t)=0\quad \text{if} \quad 0\le t\le t_0.
We’ll now show that
\label{eq:8.7.6} \lim_{h\to0+}y_h(t)=w(t-t_0)\quad \text{if} \quad t>t_0.
Suppose t is fixed and t>t_0. From Equation \ref{eq:8.7.4},
\label{eq:8.7.7} y_h(t)={1\over h}\int_{t_0}^{t_0+h}w(t-\tau)d\tau\quad \text{if} \quad h<t-t_0.
Since
\label{eq:8.7.8} {1\over h}\int_{t_0}^{t_0+h}d\tau=1,
we can write
w(t-t_0)={1\over h}w(t-t_0)\int_{t_0}^{t_0+h}\,d\tau= {1\over h}\int_{t_0}^{t_0+h}w(t-t_0)\,d\tau.\nonumber
From this and Equation \ref{eq:8.7.7},
y_h(t)-w(t-t_0)= {1\over h}\int_{t_0}^{t_0+h}\left(w(t-\tau)-w(t-t_0)\right)\,d\tau.\nonumber
Therefore
\label{eq:8.7.9} |y_h(t)-w(t-t_0)|\le {1\over h}\int_{t_0}^{t_0+h}|w(t-\tau)-w(t-t_0)|\,d\tau.
Now let M_h be the maximum value of |w(t-\tau)-w(t-t_0)| as \tau varies over the interval [t_0,t_0+h]. (Remember that t and t_0 are fixed.) Then Equation \ref{eq:8.7.8} and Equation \ref{eq:8.7.9} imply that
\label{eq:8.7.10} |y_h(t)-w(t-t_0)|\le {1\over h}M_h\int_{t_0}^{t_0+h}\,d\tau=M_h.
But \lim_{h\to0+}M_h=0, since w is continuous. Therefore Equation \ref{eq:8.7.10} implies Equation \ref{eq:8.7.6}. This and Equation \ref{eq:8.7.5} imply Equation \ref{eq:8.7.3}.
Theorem 8.7.1 motivates the next definition.
If t_0>0, then the solution of the initial value problem
\label{eq:8.7.11} ay''+by'+cy=\delta(t-t_0), \quad y(0)=0,\quad y'(0)=0,
is defined to be
y=u(t-t_0)w(t-t_0),\nonumber
where
w={\cal L}^{-1}\left(1\over as^2+bs+c\right).\nonumber
In physical applications where the input f and the output y of a device are related by the differential equation
ay''+by'+cy=f(t),\nonumber
w is called the impulse response of the device. Note that w is the solution of the initial value problem
\label{eq:8.7.12} aw''+bw'+cw=0, \quad w(0)=0,\quad w'(0)=1/a,
as can be seen by using the Laplace transform to solve this problem. (Verify.) On the other hand, we can solve Equation \ref{eq:8.7.12} by the methods of Section 5.2 and show that w is defined on (-\infty,\infty) by
\label{eq:8.7.13} w={e^{r_2t}-e^{r_1t}\over a(r_2-r_1)},\quad w={1\over a}te^{r_1t}, \quad \text{or} \quad w={1\over a\omega}e^{\lambda t}\sin\omega t,
depending upon whether the polynomial p(r)=ar^2+br+c has distinct real zeros r_1 and r_2, a repeated zero r_1, or complex conjugate zeros \lambda\pm i\omega. (In most physical applications, the zeros of the characteristic polynomial have negative real parts, so \lim_{t\to\infty}w(t)=0.) This means that y=u(t-t_0)w(t-t_0) is defined on (-\infty,\infty) and has the following properties:
y(t)=0,\quad t<t_0,\nonumber
ay''+by'+cy=0\quad \text{on} \quad (-\infty,t_0)\quad \text{and} \quad (t_0,\infty),\nonumber
and
\label{eq:8.7.14} y'_-(t_0)=0, \quad y'_+(t_0)=1/a
(remember that y'_-(t_0) and y'_+(t_0) are derivatives from the right and left, respectively) and y'(t_0) does not exist. Thus, even though we defined y=u(t-t_0)w(t-t_0) to be the solution of Equation \ref{eq:8.7.11}, this function doesn’t satisfy the differential equation in Equation \ref{eq:8.7.11} at t_0, since it isn’t differentiable there; in fact Equation \ref{eq:8.7.14} indicates that an impulse causes a jump discontinuity in velocity. (To see that this is reasonable, think of what happens when you hit a ball with a bat.) This means that the initial value problem Equation \ref{eq:8.7.11} doesn’t make sense if t_0=0, since y'(0) doesn’t exist in this case. However y=u(t)w(t) can be defined to be the solution of the modified initial value problem
ay''+by'+cy=\delta(t), \quad y(0)=0,\quad y'_-(0)=0,\nonumber
where the condition on the derivative at t=0 has been replaced by a condition on the derivative from the left.
Figure 8.7.2 illustrates Theorem 8.7.1 for the case where the impulse response w is the first expression in Equation \ref{eq:8.7.13} and r_1 and r_2 are distinct and both negative. The solid curve in the figure is the graph of w. The dashed curves are solutions of Equation \ref{eq:8.7.1} for various values of h. As h decreases the graph of y_h moves to the left toward the graph of w.
Find the solution of the initial value problem
\label{eq:8.7.15} y''-2y'+y=\delta(t-t_0), \quad y(0)=0,\quad y'(0)=0,
where t_0>0. Then interpret the solution for the case where t_0=0.
Solution
Here
w={\cal L}^{-1}\left(1\over s^2-2s+1\right)={\cal L}^{-1}\left( 1\over(s-1)^2\right)=te^{-t},\nonumber
so Theorem 8.7.2 yields
y=u(t-t_0)(t-t_0)e^{-(t-t_0)}\nonumber
as the solution of Equation \ref{eq:8.7.15} if t_0>0. If t_0=0, then Equation \ref{eq:8.7.15} doesn’t have a solution; however, y=u(t)te^{-t} (which we would usually write simply as y=te^{-t}) is the solution of the modified initial value problem
y''-2y'+y=\delta(t), \quad y(0)=0,\quad y_-'(0)=0.\nonumber
The graph of y=u(t-t_0)(t-t_0)e^{-(t-t_0)} is shown in Figure 8.7.3
Definition 8.7.2 and the principle of superposition motivate the next definition.
Suppose \alpha is a nonzero constant and f is piecewise continuous on [0,\infty). If t_0>0, then the solution of the initial value problem
ay''+by'+cy=f(t)+\alpha\delta(t-t_0), \quad y(0)=k_0,\quad y'(0)=k_1\nonumber
is defined to be
y(t)=\hat y(t)+\alpha u(t-t_0)w(t-t_0),\nonumber
where \hat y is the solution of
ay''+by'+cy=f(t), \quad y(0)=k_0,\quad y'(0)=k_1.\nonumber
This definition also applies if t_0=0, provided that the initial condition y'(0)=k_1 is replaced by y_-'(0)=k_1.
Solve the initial value problem
\label{eq:8.7.16} y''+6y'+5y=3e^{-2t}+2\delta(t-1),\quad y(0)=-3,\quad y'(0)=2.
Solution
We leave it to you to show that the solution of
y''+6y'+5y=3e^{-2t}, \quad y(0)=-3,\; y'(0)=2\nonumber
is
\hat y=-e^{-2t}+{1\over2}e^{-5t}-{5\over2}e^{-t}.\nonumber
Since
\begin{array} {ccccc}{w(t)}&{=}&{\cal{L}^{-1}\left(\frac{1}{s^{2}+6s+5} \right)}&{=}&{\cal{L}^{-1}\left(\frac{1}{(s+1)(s+5)} \right)} \\[4pt] {}&{=}&{\frac{1}{4}\cal{L}^{-1}\left(\frac{1}{s+1}-\frac{1}{s+5} \right) }&{=}&{\frac{e^{-t}-e^{-5t}}{4}} \end{array}\nonumber
the solution of Equation \ref{eq:8.7.16} is
\label{eq:8.7.17} y=-e^{-2t}+{1\over2}e^{-5t}-{5\over2}e^{-t} +u(t-1){e^{-(t-1)}-e^{-5(t-1)}\over2}
(Figure 8.7.4 ).
Definition 8.7.3 can be extended in the obvious way to cover the case where the forcing function contains more than one impulse.
Solve the initial value problem
\label{eq:8.7.18} y''+y=1+2\delta(t-\pi)-3\delta(t-2\pi), \quad y(0)=-1,\; y'(0)=2.
Solution
We leave it to you to show that
\hat y= 1-2\cos t+2\sin t\nonumber
is the solution of
y''+y=1, \quad y(0)=-1,\quad y'(0)=2.\nonumber
Since
w={\cal L}^{-1}\left(1\over s^2+1\right)=\sin t,\nonumber
the solution of Equation \ref{eq:8.7.18} is
\begin{aligned} y&=1-2\cos t+2\sin t+2u(t-\pi)\sin(t-\pi)-3u(t-2\pi)\sin(t-2\pi)\\[4pt] &=1-2\cos t+2\sin t-2u(t-\pi)\sin t-3u(t-2\pi)\sin t,\end{aligned}\nonumber
or
\label{eq:8.7.19} y=\left\{\begin{array}{cl} 1-2\cos t+2\sin t,&0\le t<\pi,\\[4pt] 1-2\cos t,&\pi\le t<2\pi,\\[4pt] 1-2\cos t-3\sin t,&t\ge 2\pi\end{array}\right.
(Figure 8.7.5 ).